View Single Post
  #3  
Old January 13th, 2009, 05:03 PM
NonCommAlg's Avatar
NonCommAlg NonCommAlg is offline
The Lord of The Rings!

 
Join Date: May 2008
Location: Vancouver, Canada
Posts: 1,995
Country:
Thanks: 237
Thanked 1,442 Times in 1,078 Posts
NonCommAlg has a brilliant futureNonCommAlg has a brilliant futureNonCommAlg has a brilliant futureNonCommAlg has a brilliant futureNonCommAlg has a brilliant futureNonCommAlg has a brilliant futureNonCommAlg has a brilliant futureNonCommAlg has a brilliant futureNonCommAlg has a brilliant futureNonCommAlg has a brilliant futureNonCommAlg has a brilliant future
Default

Quote:
Originally Posted by prometheos View Post
We just started class and our first HW problem over a chapter we haven't gotten to yet has me stumped.

The instructor made this one up I believe, and the following is from the chalkboard;

[1 2 0 1 3 | 4 ] is the reduced row echelon form if [A.b] for the eqn.
[0 0 1 2 4 | -3] Ax=b. Find solution for x.
[0 0 0 0 0 | 0 ]

From what I can gather by reading ahead in the book is for an Ax=b situation the solution is given by x=A^-1 * b... or x equals the inverse of A times b. From what I have read, inverses can only be calculated from matrices that are square or nxn dimensions. Therefore, I am stumped.

Any help is greatly appreciated, even if you just show me a general method, so I can solve it on my own.
just write Ax=b as a system of equations and solve it: \begin{cases} x_1 + 2x_2 + x_4 + 3x_5=4 \\ x_3 + 2x_4 + 4x_5 =-3 \end{cases}. we have two equations and 5 variables.

the solutions are: x_1=4-2x_2-x_4-x_5, \ x_3=-3-2x_4-4x_5. note that x_2, x_4, x_5 are free variables.
Reply With Quote